11th Class Chemistry Redox Reactions

  • question_answer 29) Balance the following equations in basic medium by ion-electron method and oxidation number method : (a) \[{{P}_{4}}(s)+O{{H}^{-}}(aq)\to P{{H}_{3}}(g)HPO_{2}^{-}(aq)\] (b) \[{{N}_{2}}{{H}_{4}}(l)+ClO_{3}^{-}(aq)\to P{{H}_{3}}(g)+HPO_{2}^{-}(aq)\] (c) \[{{N}_{2}}{{H}_{4}}(l)+ClO_{3}^{-}(aq)\to P{{H}_{3}}(g)+HPO_{2}^{-}(aq)\] \[+{{O}_{2}}(g)+{{H}^{+}}\]  

    Answer:

    (a) Given equation is : \[{{P}_{4}}+O{{H}^{-}}\to P{{H}_{3}}+HPO_{2}^{-}\] half equations are : \[{{P}_{4}}\to P{{H}_{3}}\] \[{{P}_{4}}\to HPO_{2}^{-}\] These half equations can be balanced as \[[{{P}_{4}}+12{{H}^{+}}+12{{e}^{-}}\to 4P{{H}_{3}}]\times 2\] \[On\,addition\frac{\frac{[{{P}_{4}}+8{{H}_{2}}O\to 4HPO_{2}^{-}+12{{H}^{+}}+8{{e}^{-}}]\times 3}{5{{P}_{4}}+24{{H}_{2}}O\to 8P{{H}_{3}}+12HPO_{2}^{-}+12{{H}^{+}}}}{{}}\]converting above equation into basic medium we get \[5{{P}_{4}}+12{{H}_{2}}O+12O{{H}^{-}}\to 8P{{H}_{3}}+12HPO_{2}^{-}\] (b) Given equation is : \[{{N}_{2}}{{H}_{4}}(l)+ClO_{3}^{-}(aq)\to NO(q)+C{{l}^{-}}(aq)\] Half equations are: \[{{N}_{2}}{{H}_{4}}(l)\to NO(q)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.......(i)\] \[{{N}_{2}}{{H}_{4}}(l)\to NO(q)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.......(i)\] These half equations can be balanced as \[[{{N}_{2}}{{H}_{4}}(l)]+2{{H}_{2}}O\to 2NO+8{{H}^{+}}+8{{e}^{-}}]\times 3\] On \[ONaddition\frac{[ClO_{3}^{-}(aq)]+6{{H}^{+}}+6{{e}^{-}}\to C{{l}^{-}}(aq)+3{{H}_{2}}O]\times 4}{\frac{3{{N}_{2}}{{H}_{4}}(l)+4ClO_{3}^{-}(aq)\to 6NO+4C{{l}^{-}}(aq)+6{{H}_{2}}O}{{}}}\](c) \[C{{l}_{2}}{{O}_{7}}(g)+{{H}_{2}}{{O}_{2}}(aq)\to ClO_{2}^{-}(aq)\] \[+{{O}_{2}}(g)+{{H}^{+}}(aq)\] Step I: Splitting the equation into two half equations: \[C{{l}_{2}}{{O}_{7}}(g)\to ClO_{2}^{-};\] \[{{H}_{2}}{{O}_{2}}(aq)\to {{O}_{2}}(g)+{{H}^{+}}(aq)\] Step II: Balancing the element other than hydrogen and oxygen: \[C{{l}_{2}}{{O}_{7}}(g)\to ClO_{2}^{-};\] \[{{H}_{2}}{{O}_{2}}(aq)\to {{O}_{2}}(g)+{{H}^{+}}(aq)\] Step III: Balancing oxygen by adding water molecules: \[C{{l}_{2}}{{O}_{7}}(g)\to ClO_{2}^{-}+3{{H}_{2}}O\] \[{{H}_{2}}{{O}_{2}}(aq)\to {{O}_{2}}(g)+{{H}^{+}}(aq)\] Step IV: Balancing hydrogen by adding \[{{H}^{+}}\] ions. \[C{{l}_{2}}{{O}_{7}}+6{{H}^{+}}+8{{e}^{-}}\to 2ClO_{2}^{-}+3{{H}_{2}}O\] \[{{H}_{2}}{{O}_{2}}(aq)\to {{O}_{2}}(g)+2{{H}^{+}}\] Step V: Balancing charge by adding electron, we get \[C{{l}_{2}}{{O}_{7}}(g)+6{{H}^{+}}+8{{e}^{-}}\to 2ClO_{2}^{-}+3{{H}_{2}}O\] \[[{{H}_{2}}O(aq)\to {{O}_{2}}(g)+2{{H}^{+}}+2{{e}^{-}}]\times 4\] Step VI: Adding above equations, we get \[C{{l}_{2}}{{O}_{7}}(g)+4{{H}_{2}}{{O}_{2}}(aq)\to 2ClO_{2}^{-}+3{{H}_{2}}O+2{{H}^{+}}\] This equation can be balanced in basic medium by adding two \[O{{H}^{-}}\] ions on both sides: \[C{{l}_{2}}{{O}_{7}}(g)+4{{H}_{2}}{{O}_{2}}(aq)2O{{H}^{-}}\to 2ClO_{2}^{-}+\] \[3{{H}_{2}}O+2{{H}^{+}}+2O{{H}^{-}}\] OR \[C{{l}_{2}}{{O}_{7}}(g)+4{{H}_{2}}{{O}_{2}}(aq)2O{{H}^{-}}(aq)\to 2ClO_{2}^{-}+5{{H}_{2}}O\]  


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